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+// Theory and Problems of Thermodynamics

+// Chapter 9

+// Air_water Vapor Mixtures

+// Example 2

+

+clear ;clc;

+

+//Given data

+X1 = 0.5                // volume fraction of propane 

+X2 = 0.5                // volume fraction of oxygen 

+P1 = 0.1                // initial pressure of mixture in MPa 

+T1 = 300                // initial temperature of mixture in K

+P2 = 1.0                // final pressure of mixture in MPa

+N = 100                 // number moles of gas mixture

+Cp1 = 74.0565           // molar heat capacity of propone in kJ/mol K

+Cv1 = 65.7442           // molar heat capacity of propone in kJ/mol K

+Cp2 = 29.4912           // molar heat capacity of oxygen  in kJ/mol K

+Cv2 = 21.1776           // molar heat capacity of oxygen  in kJ/mol K

+

+// Calculations

+Cp_m = X1*Cp1 + X2*Cp2  // molar heat capacity of gas mixture

+Cv_m = X1*Cv1 + X2*Cv2  // molar heat capacity of gas mixture

+

+gam = Cp_m/Cv_m         // specific heat ratio

+

+T2 = T1*((P2/P1)^((gam-1)/gam))  // final temperature of mixture

+

+W = -N*Cv_m*(T2-T1)              // Work  required to compress gas mixture

+

+W = W*1e-3                       // units conversion J to kJ

+

+// Output Results

+mprintf('Final temperature of gas mixture = %4.1f K' ,T2);

+mprintf('\n Work  required to compress 100 mol of gas mixture = %4.2f kJ' ,W);